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Math 131
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Fall 2012
Problem 1. Let f ( x) = x + ln(1 + x) .
a. Show that the graph of f is always concave down.
b. Find the linear approximation for f ( x) = x + ln(1 + x) near x = 0 .
c. Use your linear approximation from Part b to estimate f (0.5) .
d. Is your answer to Part c an overestimate or underestimate? Explain.
e. Find a bound on the error for your approximation.
f. Use your linear approximation from Part b to approximate where f ( x) = 0.2
g. Is your answer to Part f an overestimate or underestimate? Explain.
Solution.
a.
b.
−1
1
so f ′′( x) =
. Since (1 + x)2 > 0 for all x > −1 , then f ′′( x) < 0 for the
2
1+ x
(1 + x)
entire domain of f.
f ′( x) = 1 +
f (0) = 0 and f ′(0) = 2 so the tangent line passes through (0,0) with slope 2. Thus the
equation of the tangent line is y = 2 x .
c. Using f ( x) ≈ 2 x , we get f (0.5) ≈ 1 .
d. The estimate in Part c is an overestimate because the graph of f is concave down, which
means the tangent line lies above the graph.
e. Using the error bound formula for a tangent line approximation,
E (0.5) <
M
2
0.5 where M = max f ′′ between x = 0 and x = 0.5 .
2
1
, its maximum on the relevant interval is when x is smallest ( x = 0 )
(1 + x) 2
so M = 1 . Thus
Since f ′′( x ) =
E (0.5) <
1
1
2
0.5 = .
2
8
f. We can approximate where f ( x) = 0.2 by finding where the tangent line y = 2 x = 0.2 is
zero. Specifically, we solve for x to obtain x = 0.1 .
g. The approximation x = 0.1 for where f ( x) = 0.2 is an underestimate. Since f ′′( x) < 0 , the
graph of f is concave down, so the tangent line will be above the graph of f and thus will
reach the height y = 0.2 before the actual function does.
Problem 2. Let U and V be positive constants and F (t ) = Uet + Ve − t .
a. Find all critical points of F.
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
b. Use the first derivative to determine if your critical points result in local minima, local
maxima, or neither.
Solution.
a. F ′(t ) = Uet − Ve− t so the critical points are when Ue t − Ve − t = 0 . Solving for t we get
V
V
e 2t =
or t = 12 ln .
U
U
b. We need to know whether F ′(t ) = Uet − Ve− t is positive or negative to the left and right of
V
the critical point t = 12 ln . We can determine this by noticing that as t → −∞ , the
U
exponential in the first term goes to 0 while the exponential in the second term goes to ∞ .
V
Thus F ′(t ) = Uet − Ve −t < 0 when t < 12 ln . Similarly as t → ∞ , the exponential in the first
U
term goes to ∞ while the exponential in the second term goes to 0. Thus
V
F ′(t ) = Uet − Ve −t > 0 when t > 12 ln . This means F is decreasing to the left of the critical
U
V
point and increasing to the right, so F has a local minima at t = 12 ln .
U
Problem 3. A warehouse selling cement has to decide how often and in what quantities to
reorder. It is cheaper, on average, to place a smaller number of larger orders. On the other hand,
larger orders mean higher storage costs. The warehouse always reorders cement in the same
quantity, q. The total weekly cost, C, of ordering and storage is given by
C=
a
+ bq
q
where a and b are positive constants.
a. Which of the terms a q or bq, represents the ordering cost and which represents the storage
costs? Explain.
b. What value of q gives the minimum total cost?
Solution.
a. a q represents the weekly cost of ordering since it is a decreasing function of q and placing
larger orders (larger q) reduces the weekly ordering cost. On the other hand bq is an
increasing function of q and placing larger orders increases the weekly storage cost.
b. First note that the practical domain of C as a function of q is 0 < q < ∞ .
is undefined when q = 0 and zero when q =
dC −a
=
+ b which
dq q 2
a
. Since q = 0 is not in the domain of C, we
b
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
only have one critical point. For values of q near 0,
dC
< 0 and for large values of q,
dq
dC
a
> 0 . Thus C has a global minimum at q =
.
dq
b
Problem 4. For which positive number x is x1/ x largest? Justify your answer.
Hint: Take the log of both sides y = x1/ x so you can move the 1/x out of the exponent. Then use
implicit differentiation.
Solution. Using the hint, we first simplify y = x1/ x so ln y = ln x1/ x = 1x ln x . Using implicit
differentiation,
1 dy x 1x − ln x 1 − ln x
dy x 1x − ln x 1 − ln x 1 x
=
=
.
So
=
=
x . This is undefined at
y dx
x2
x2
dx
x2
x2
x = 0 , but that is not in the domain of the original function. So the only critical point is when the
derivative is zero. This happens when 1 − ln x = 0 which is at x = e .
Problem 5. A rectangular swimming pool is to be built with an area of 1800 square feet. The
owner wants 5-foot wide decks along either side and 10-foot wide decks at the two ends. Find
the dimensions of the smallest piece of property on which the pool can be built satisfying these
conditions.
Solution. We sketch a generic picture of the configuration of the pool and decks:
Deck
10 ft.
5 ft.
Pool
(Area = 1800 ft.2)
y
x
The area of the pools is xy = 1800 ft.2 and we want to minimize the entire deck+pool area which
is A = ( x + 20)( y + 10) . So we can rewrite this as a function of only x as
(
)
A( x) = ( x + 20) 1800 + 10 = 2000 + 10 x + 36000 . The practical domain of this function is
x
x
0 < x < ∞ . Taking the derivative, we get A′( x) = 10 − 36000 2 which does not exist at x = 0 and
x
is zero when x = 60 . Since x = 0 is not in the domain and there are no endpoints, the local
extrema occurs at x = 60 . Plug = 60 into = 1800ft to get = 30. We can verify it is a
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
minimum by noting that if x < 60 then A′( x) < 0 but if x > 60 then A′( x) > 0 . Thus the smallest
piece of property for the pool and deck is 80 ft by 40 ft.
Problem 6. The marketing department of computer giant, Pear, determines that the revenue (in
billions of dollars) generated by selling q million units of their popular pPods is given by
R (q ) = 0.8ln(1 + 2.5q 2 )
while the cost (also in billions of dollars) is determined by
C (q) = 0.7 + 0.5q
Find the number of pPods that Pear should produce to maximize their profit. What is their profit
in this case?
Solution. The profit is P (q ) = R (q ) − C (q ) = 0.8ln(1 + 2.5q 2 ) − (2 + 0.5q ) with domain 0 ≤ q < ∞ .
The derivative is P ′( q ) =
0.8(5q )
− (1.25q 2 − 4 q + 0.5)
which is zero when
−
0.5
=
1 + 2.5q 2
1 + 2.5q 2
4 ± 42 − 4(1.25)(.5)
= 1.6 ± 1.470 . We can find the global maximum by checking the
2(1.25)
endpoint, q = 0 , and the two critical points, q = 0.13 and q = 3.07 :
q=
P (0) = −0.7 , P (0.13) = −0.732 , and P (3.07) = 0.956 .
Thus Pear should sell 3,070,000 pPods which will generate a profit of $956,000,000.
Problem 7. Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the
gasoline is 4 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline
changing at that instant?
Solution. The volume and depth of gasoline in the tank are related by the equation
= ℎ = 9ℎ. Treating V and h as functions of time, we differentiate to get
Since
dV
dh
= 9π
.
dt
dt
dh
dV
= 0.2 ft/s, we can solve for
= 9π ⋅ 0.2 = 5.655 ft3/s.
dt
dt
Problem 8. Suppose a rectangular beam is cut from a cylindrical log of radius 15 cm. The
strength of a beam is proportional to the product of the width and the square of the height. Find
the width and height of the beam with maximum strength that can be cut from the log. Fully
justify your answer.
Solution. First we draw a diagram
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
w
Since the radius of the log is 15 cm, the diagonal of a beam cut from it will be 30 cm. Labeling
the width as w, we can use the Pythagorean theorem to find the height of the beam. Then the
strength of the beam is
S ( w) = kw ( 900 − w2 ) = 900kw − kw3 ,
with domain 0 < w < 30 . The derivative of this function is
S ′(w) = 900k − 3kw2 .
So S has critical points when
3kw2 = 900k
w2 = 300
w = 10 3.
As w → 0 or as w → 30 , either the width or the height of the beam goes to 0, thus S (w) → 0 .
Since S (w) > 0 for all w in the domain, the critical point must correspond to a maximum. Thus
the width and height of the strongest beam that can be cut from the log are 10 3 ≈ 17.3 cm and
900 − 300 ≈ 24.5 cm, respectively.
Problem 9. Representative values of the derivative of a function f ( x) are shown in the table
below. Assume f ( x) is a continuous function and that the values in the table are representative
of the behavior of f ′( x) .
x
0
0.5
1
1.5
2
2.5
3
f ′( x)
1
0.3
0
–0.1
–0.15
–0.12
–0.1
a. Estimate the x-value of the global maximum and minimum of f ( x) on the closed interval
[0,3] . Justify your answers based on the data in the table.
b. Can you tell from these data if f ( x) has any inflection points? If so, estimate the x-value of
any inflection points and indicate how you know their locations. If not, explain why not.
Solution.
Math 131
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Fall 2012
a. We note that f ′( x) > 0 for x < 1 and f ′( x) < 0 for x > 1 . Thus f ( x) has a local maximum
at x = 1 . Further, because there is only one change of sign in the derivative, we know that
this is the global maximum. The global minimum will occur at one of the endpoints. It is not
easy to tell at which endpoint this occurs, but because the negative slopes are of smaller
magnitude (for x > 1 ) than the positive slopes (for x < 1), we expect that the global minimum
occurs at x = 0 .
b. We know that an inflection point occurs when f ′( x) goes from increasing to decreasing or
vice versa. We can see from these data that f ′( x) is decreasing until sometime between
x = 2 and x = 2.5 , and increasing thereafter. Thus there is an inflection point at an x
somewhere in (2, 2.5).
Problem 10. Recall that arccosine is the inverse of cosine, so that cos(arccos( x)) = x .
d
arccos x . Express your final
dx
answer in a form that does not depend on any trigonometric functions.
a. Differentiate both sides of this equation then solve for
b. Use the local linearization of arccos x near x = 0 to approximate arccos ( 12 ) .
Hint: You will need to use the value of arccos ( 0 ) .
c. Is your answer to part b an overestimate or underestimate? Explain your answer graphically.
Hint: You will need to describe the shape of the graph of y = arccos ( x ) .
Solution.
a. Differentiating both sides of the equation cos(arccos( x)) = x yields
− sin(arccos( x))
d
(arccos( x)) = 1 ,
dx
and solving for the derivative, we get
d
1
(arccos( x)) =
.
dx
− sin(arccos( x))
We can simplify this by applying the Pythagorean theorem to a right-triangle with angle
arccos( x) :
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
1
1 − x2
arccos( x)
x
From this diagram, we can read off the sin(arccos( x)) = 1 − x2 . Thus
d
1
.
(arccos( x)) =
dx
− 1 − x2
b. At x = 0 , the derivative of arccos( x) is
1
− 1 − 02
= −1 . Since arccos(0) =
π
2
, the tangent line
 π
passes through  0,  with slope –1:
 2
arccos( x) ≈
π
2
−x.
Evaluating this at x = 12 ,
arccos ( 12 ) ≈
c. Observe
π
2
−
1
≈ 1.0708 .
2
arcos = −
1 − / after simplification. Observe that for 0 < < 1 the second derivative is negative which
means that the graph of y = arccos( x) , is concave down for 0 < x < 1 , which implies the
tangent line will lie above the graph in this region. Thus our approximation is an
overestimate.
2
Problem 11. Consider the function f ( x) = xe− x . Find and classify all critical points of this
function as minima, maxima or neither, and use this information to sketch the graph of f ( x) .
Where is f ( x) increasing? Decreasing? On your graph, mark the approximate location of the
inflection points (you do not need to find them algebraically). How many inflection points are
there? Does this function have an absolute minimum? Absolute maximum? Find the absolute
minimum and absolute maximum of f ( x) on the interval [−2,3] . Where are they attained?
Solution. The domain of f is all real numbers, −∞ < x < ∞ , so any place where the derivative is
zero or undefined is a critical point. The derivative is
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
2
f ′( x) = e− x − 2 x 2 e− x
2
= e− x (1 − 2 x 2 )
2
2
which is continuous for all x. Since e− x is never zero, the critical points are when 1 − 2 x2 = 0 ,
that is, x = ± 12 . Evaluating the derivative at x = −1 , x = 0 , and x = 1 will then give us
complete information on where f ′ is positive and where it is negative (since f ′ is continuous
and can only switch sign at its zeroes):
f ′(−1) = −e−1 < 0 , f ′(0) = 1 > 0 , and f ′(1) = −e−1 < 0 .
(
Thus f is decreasing on −∞,
at
−1
2
and a local maximum at
−1
2
)∪(
1
2
1
2
)
, ∞ and increasing on
(
−1
2
,
1
2
) and has a local minimum
. Also note that since f ( x) → 0 as x → ±∞ , these are in fact
the global minimum and maximum for f. Specifically, f
( ) = −0.4289 and f ( ) = 0.4289 .
−1
2
1
2
Using this information, we can plot the graph of f:
There are three inflection points (where the graph of f changes concavity) marked on the graph.
Although the problem only asked us to estimate the locations of the inflections points on the
2
graph, we can locate them precisely from f ′′( x) = −2 x 3 − 2 x 2 e− x which is zero when x = 0
(
)
and x = ± 32 . Finally, since the interval [−2,3] contains the global minimum and maximum for
f, these are also the absolute extrema for the interval.
Problem 12. The hypotenuse of a right triangle has one end at the origin and one end on the
Curve y = x 2e−3x , with x ≥ 0 . One of the other two sides is on the x-axis, the other side is
parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?
Math 131
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Fall 2012
Solution. We are looking for the maximum value of the area of a triangle with base x and height
y = x 2e−3x . That is, we want to maximize A( x) = 12 x3e −3 x for values of x ≥ 0 . The derivative is
A′( x) = 32 x 2 e−3 x − 32 x3e −3 x = 32 x 2 (1 − x)e −3 x , which is zero when x = 0 and x = 1 . The only factor
in this expression that can be negative is the (1 − x) factor, so A′( x) is positive just to the left of
1 and negative just to the right. Thus A has a maximum at x = 1 . The maximum area is then
A(1) = 12 e −3 ≈ 0.0249 .
y
x
y = x 2 e −3 x
Problem 13. A model for the amount of an antihistamine in the bloodstream after a patient takes
a dose of the drug gives the amount, a, as a function of time, t, to be a (t ) = A ( e − t − e − kt ) . In this
equation, A is a measure of the dose of antihistamine given to the patient, and k is a transfer rate
between the gastrointestinal tract and the bloodstream. A and k are positive constants, and for
pharmaceuticals like antihistamine, k > 1 .
a. Find the location t = Tm of the non-zero critical point of a(t ) .
b. Explain why t = Tm is a global maximum of a(t ) by referring to the expression for a(t ) or
a′(t ) .
c. The function a(t ) has a single inflection point. Find the location t = TI of this inflection
point. You do not need to prove that this is an inflection point.
d. Using your expression for Tm from Part a, find the rate at which Tm changes as k changes.
Solution.
a. The maximum will occur at an endpoint or at a critical point, when a′(t ) = 0 . The critical
points are thus where a′(t ) = A ( −e− t + ke − kt ) = 0 . Solving, we have e−t = ke−kt , so that
e( k −1)t = k , or t = Tm = k1−1 ln k .
Math 131
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Fall 2012
b. Note that a′(0) = A(k − 1) > 0 and that for large t, a′(t ) = A ( −e− t + ke − kt ) < 0 since k > 1
guarantees that the second exponential decays much faster than the first. Thus the critical
point must be a maximum. In addition, because t = Tm this is the only critical point so we
know it must be the global maximum.
Alternately, note that a(0) = 0 . Because k > 0 , a(t ) ≥ 0 for all t (the exponential involving
−kt will decay faster than e − t ). And for large t, a(t ) → 0 . Thus at t = Tm , a(t ) must take on
a maximum value, and because it is the only critical point this must be the global maximum.
c. To find inflection points, we look for where a′′(t ) = A ( e− t − k 2 e − kt ) = 0 . Solving for t, we
have e( k −1)t = k 2 , so that t = TI =
1
k −1
ln k 2 =
2
k −1
ln k .
Although the problem did not ask, we could show that this is an inflection point by a similar
argument to (b): because k > 1 we know that a′′(0) < 0 and as t increases a′′(t ) must
eventually become positive. Thus this is an inflection point.
d. We have Tm =
1
k −1
ln k . Thus
dTm
1
1
=−
ln k +
.
2
(k − 1)
dk
k (k − 1)
Problem 14. If C (q) = 0.02q 2 − 1.2q + 28 is the total cost to produce q lightsabers and
R(q) = 48q is the revenue from selling q light sabers then:
a. Find the average cost to produce q lightsabers.
b. Find the average revenue from selling q lightsabers.
c. Identify the fixed cost and the variable cost.
d. Find the maximum profit.
e. Find and interpret C′(28) in the context of the problem.
Solution.
a. The average cost to produce q ligthsabers is
C (q )
28
= 0.02q − 1.2 +
dollars per lightsaber.
q
q
b. The average revenue from selling q ligthsabers is
R (q)
= 48 dollars per lightsaber.
q
c. The fixed cost is 28 dollars. The variable cost is 0.02q 2 − 1.2q dollars.
d. The profit is P(q) = R(q) − C (q) = −0.02q 2 + 49.2q − 28 so P′(q) = −0.04q + 49.2 which is
zero when q = 1230 . Since P′(1000) = 9.2 and P′(1250) = −0.8 , C is increasing to the left of
the critical point q = 1230 and decreasing to the right. Thus C(1230) = $30, 230 is the
maximum profit.
Math 131
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CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
e. C′(28) = 0.04 ⋅ 28 − 1.2 = −0.08 . This means that when producing 28 lightsabers, for some
strange reason, it will actually cost about 8 cents less to produce one more.
Problem 15. For the following function:
f ( x) = 2 x3 + 3x2 − 12 x − 5
NO CREDIT will be given if there is no work shown.
a. Find all critical value(s).
b. Classify all critical values from Part a. as local maxima, local minima, or neither.
c. Find all inflection points.
d. Find the global maxima and minima over the closed interval [−3,3] .
e. Sketch the graph labeling all of the points found in the previous parts.
Solution.
a.
f ′( x) = 6 x 2 + 6 x − 12 = 6( x + 2)( x − 1) so the critical points are at x = −2 and x = 1 .
b. If x < −2 , both factors in the derivative will be negative, so f ′( x) > 0 .
If −2 < x < 1 , the x + 2 factor will be positive but x − 1 will be negative, so f ′( x) < 0 .
If x > 1 , both factors in the derivative will be positive, so f ′( x) > 0 .
This means that f will have a local maximum at x = −2 and a local minimum at x = 1 .
c.
f ′′( x) = 12 x + 6 , and if x < − 12 then f ′′( x) < 0 , but if x > − 12 then f ′′( x) > 0 . Thus the
concavity of f switches from concave down to concave up, so x = − 12 is an inflection point.
d. Evaluating f at both critical points and both endpoints, f (−3) = 4 , f (−2) = 15 , f (1) = −12
and f (3) = 40 . Thus over the closed interval [−3,3] , f has a global maximum of 40 at x = 3
and a global minimum of –12 at x = 1 .
Problem 16. Find the following limits.
3
x+7 −2
a. lim
x→1
x −1
1
 x + 2 x
b. lim 

x →0  2 
x2
c. lim − x
x →0 5 + (ln 5) x − 1
Solution.
3
a. lim
x→1
x+7 −2 "0"
=
so we can use L’Hopital’s Rule.
x −1
0
1 x+7
(
)
x+7 −2
lim
= lim 3
x→1
x→1
x −1
1
3
−2 3
= 13 8−2 3 =
1
12
1
1
d. lim − t
t →0 t e − 1
Math 131
Fall 2012
CHAPTER 4 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
1
1
 x + 2 x
 x + 2 x
= "1∞ " so we will need to rewrite this by setting y = 
b. lim 

 then
x →0  2 
 2 
1  x+2
ln y = ln 
 . Now,
x  2 
ln x +2 2
"0"
lim ln y = lim
=
so we can now apply L’Hopital’s Rule:
x →0
x →0
x
0
( )
lim ln y = lim
x →0
x →0
12
So lim y = e
x →0
c.
ln
( x+2 2 ) = lim 12 x+2 2 = 1 .
x →0
x
x →0 5 − x
=
+ (ln 5) x − 1
x2
lim
x →0 5− x
lim
x→0 − (ln 5)5
"0"
0
so we can use L’Hopital’s Rule:
2x
= lim
x →0 −(ln 5)5− x
+ (ln 5) x − 1
2x
−x
2
.
x2
lim
1
+ (ln 5)
+ (ln 5)
2
= lim
x→0 (ln 5)
2 −x
5
=
=
"0"
0
so we can use L’Hopital’s Rule again:
2
(ln 5)2
1
1
= " ∞ − ∞ " so we need to rewrite the function as a single fraction,
d. lim − t
t →0 t e − 1
lim
et − 1 − t
t →0 t
lim
et − 1 − t
t →0 t
lim
( et − 1)
(
t → 0 et
)
et − 1
=
"0"
0
= lim
t →0 et
et − 1
− 1 + te
so we can now apply L’Hopital’s Rule:
t
= lim
et − 1
− 1 + te
t →0 e t
t
=
"0"
0
et
t
+ e + te
t
=
so we can use L’Hopital’s Rule again:
1
.
2